Practice Test: Quantitative Aptitude - 2 - SSC CGL MCQ

Consider the trapezium ABCD,
where AB = 6 cm, BC = 12 cm, CD = 8 cm and AD = 22 cm.
Draw a line from C parallel to AB at E such that ABCE is a parallelogram.
∴ CE = AB = 6 cm and AE = BC = 12 cm.
AD = AE + ED
⇒ ED = AD – AE
⇒ ED = 22 – 12 = 10 cm.
Area of Trapezium ABCD = Area of Parallelogram ABCE + Area of ΔCED
Draw a line through C to ED such that CF is perpendicular to ED.
Area of ΔCED:
From Heron’s formulae,
Δ CED = √[(s) (s - a) (s - b) (s - c)], where s = (a + b + c)/2 and a, b, c are side of the triangle.
In this case, a = 6; b = 8 and c = 10
⇒ s = 24/2 = 12
Area of ΔCED = √(12 × 6 × 4 × 2)
⇒ Area of ΔCED = √(576) = 24 sq. cm.
Area of ΔCED = (1/2) × ED × CF
⇒ 24 = (1/2) × 10 × CF
⇒ CF = 4.8 cm.
Area of Parallelogram ABCE = AE × CF
⇒ Area of Parallelogram ABCE = 12 × 4.8 = 57.6 sq. cm.
Area of trapezium ABCD = Area of parallelogram ABCE + Area of ΔCED
⇒ Area of trapezium ABCD = 57.6 + 24 = 81.6 sq. cm.
∴ Approximate area of the trapezium = 81.6 sq. cm.

From given data,
Total investment = Rs. 6000
A investment = Rs. 1000
⇒ B and C’s investment = 6000 – 1000 = Rs. 5000
The ratio of B and C’s investment = 2 : 3
⇒ B’s investment = 2/5 × 5000 = Rs. 2000
⇒ C’s investment = 3/5 × 5000 = Rs. 3000
∴ ratio of their investment = 1000 : 2000 : 3000 = 1 : 2 : 3
Given that the annual profit = Rs. 2400
Let their profits be x : 2x : 3x
⇒ 2400 = x + 2x + 3x
⇒ 6x = 2400
⇒ x = Rs. 400
∴ C’s profit = 3x = 3 × 400 = Rs. 1200

A train 150 metres long crosses a pole in 15 seconds
Distance covered to cross the pole = Length of the train = 150 m
Time required = 15 sec
∴ Speed of the first train
= Distance covered/Required time
= 150m/15sec = 10 m/sec
Let, Speed of the second train = ‘x’ m/sec
∵ Trains were travelling in opposite directions
∴ Relative velocity of first train with respect to second train
= (10 + x) km/sec
Given,
First train takes 12 second to cross the second train moving in opposite direction
∴ Distance covered to cross second train
= Length of first train + Length of second train
∵ Length of second train = Length of first train
= 150 m + 150 m = 300 m
And, Time required = 12 sec
∴ Speed of the first train = Distance covered/Required time
⇒ 10 + x = 300/12
⇒ 120 + 12x = 300
⇒ 12x = 180

From the given data
But we know that x³ + 1/x³ = 0

Squaring on both sides, we get

Hence, On 23rd day the work will be completed.

We know that sum of the interior angles of a polygon of n sides = 180 (n - 2)
Given that sum of the measures of all the interior angles = 1080°
⇒ 180(n – 2) = 1080
⇒ n – 2 = 6
⇒ n = 8

From given data, we get
⇒ 2cosec² A = x
⇒ 2/sin² A = x

From given data
No of students failed in Hindi n(H) = 35%
No of students failed in English n(E) = 45%
No of students failed in both n(H ∩ E) = 20%
⇒ total number of students who failed n(H ⋃ E) = n(H) + n(E) – n(H ∩ E)
⇒ n(H ⋃ E) = 35 + 45 - 20
⇒ n(H ⋃ E) = 60
No of students who passed = 100 - (total no of students who failed) = 100 – 60 = 40%

To solve this problem, we need to find the interest rate and then calculate the compound interest for four years.

Let's first find the interest rate. We know that the amount doubles in two years at compound interest. So, we can use the formula for compound interest:

A = P(1 + r/n)^(nt)

Where:
A = final amount
P = principal amount (initial investment)
r = annual interest rate (decimal)
n = number of times the interest is compounded per year
t = number of years

We are given:
A = Rs. 6000
P = Rs. 3000
t = 2 years
n = 1 (interest compounded annually)

Plugging these values into the formula, we have:

6000 = 3000(1 + r/1)^(1*2)
6000 = 3000(1 + r)^2

Dividing both sides by 3000:

2 = (1 + r)^2

Taking the square root of both sides:

sqrt(2) = 1 + r

Now, solving for r:

r = sqrt(2) - 1

Now we have the interest rate. Next, let's use this interest rate to find the compound interest for four years.

Using the formula for compound interest again, this time with t = 4 years:

A = P(1 + r/n)^(nt)

A = 3000(1 + (sqrt(2) - 1)/1)^(1*4)

Calculating this expression, we get:

A = 3000(2)^2 = 3000 * 4 = Rs. 12,000

Since the initial investment was Rs. 3000, the interest earned for four years is:

Interest = A - P = 12000 - 3000 = Rs. 9000

Hence, the answer is (1) Rs. 9000.

Half of each of the two diagonal forms the two sides of a right - angled triangle whose hypotenuse is the side of the rhombus.
Length of Diagonal 1 = 24
Length of Diagonal 2 = 10
⇒ Sides of the triangle = 12 and 5

Since it’s a right - angled triangle, it should follow Pythagoras theorem.
⇒ (Side)² = (12)² + (5)²
⇒ Side = 13
∴ Side of the rhombus = 13.
⇒ Thrice the value of the side = 3 × 13 = 39

Let the angles of the triangles are in the ratio x ∶ 2x ∶ 3x
We know that sum of the angles in a triangle = 180°
⇒ x + 2x + 3x = 180
⇒ 6x = 180°
⇒ x = 30°
by sin formula sinA/a = sinB/b = sinC/c
∴ the angles of the triangle are 30°, 60° and 90°, so the sides are in the ratio of 1 ∶ 2 ∶ √3
∴ smallest side will be opposite to 30°
Let the sides of the triangle are in the ratio of x : 2x : √3x
But given smallest side x = 10 cm
⇒ remaining sides are 2x = 2(10) = 20 cm and √3x = √3(10) = 17.32 cm
∴ the longest side = 20 cm

Let the third number be X.
Let the first number be A, and the other number be B. We have,

Let the average of 6 innings be x.
Then new average can be found by (6x + 100)/7 = x + 7
⇒ 6x + 100 = 7x + 49
⇒ x = 51
∴ New average = 51 + 7 = 58

a³ – b³ = (a – b)(a² + ab + b²)
56 = 2(a² + ab + b²)
28 = (a² + ab + b²) ----(1)
Now, (a – b)² = a² – 2ab + b² = 4 ----(2)
Solving equation (1) and (2),
a² + ab + b² = 28
a² – 2ab + b² = 4
3ab = 24
ab = 8
Therefore, a² + 8 + b² = 28 in (1)
And a² + b² = 28 – 8 = 20

We know that in a triangle,
Sum of angles = 180°
⇒ ∠A + ∠B + ∠C = 180°
⇒ 3∠C + 5∠C + ∠C = 180°
⇒ 9∠C = 180°
⇒ ∠C = 180/9 = 20°
∴ ∠C = 20°

tanθ = perpendicular/base
Given, a pole stands vertically, inside a scalene triangular park ABC. The angle of elevation of top of the pole from each corner of the park is same.
If the angle of elevation is same then, the distance of each corner from the foot of the would also be same,
Base = perpendicular/tanθ
Perpendicular is height of the pole while θ is the angle of elevation.
For the above to be true, the foot of the pole should be a circumcenter and the distance of each corner from the foot is the circumradius.

From the given data
Length of sides of a triangle are of 5 cm, 9 cm and x cm
⇒ length of the third side must be less than the sum of the other two sides
⇒ x < 5 + 9 or x < 14
We also know that length of the third side must be greater than difference of the other two sides
⇒ x > 9 – 5 or x > 4
⇒ By both the conditions 4 < x < 14
∴ the answer is 5

From given data
Let AB be a chord of circle with center O and radius OA and OB equals to 13 cm
Draw OP perpendicular to AB
Given that OP = 5 cm
From ΔOPA, we know that
⇒ OA² = AP² + OP²
⇒ 13² = AP² + 5²
⇒ 169 – 25 = AP²
⇒ AP = 12 cm
But we know that AB = 2 AP = 2× 12 = 24 cm

Let the first part of 43 be x, then the second part will be 43 – x
According to the question,
4x + 3 × (43 – x) = 150
⇒ 4x + 129 – 3 x=150
⇒ x = 21
∴ The first part is 21 and second part is 43 – 21 = 22

Let the cost price of the item be Rs. 100
25% discount is offered on an item
⇒ discount = 100 × 25/100 = Rs. 25
∴ discounted selling price = 100 - 25 = Rs. 75
Again 20% cash back is offered
⇒ cash back amount = 75 × 20/100 = Rs. 15
∴ final selling price = 75 - 15 = Rs. 60
Effective discount = cost price – selling price/cost price × 100 = 100 - 60/100 × 100 = 40%

As per the given data,
3$ was the GDP at the end of 2013.
It was given that to find at the beginning of 2015, it means to find the GDP at the end of 2014. At the end of 2014, there was the increase of GDP to 8%.
⇒ total amount at the end of 2014 = 3 trillion + (8% of 3 trillion) = 3.24 trillion
∴ GDP at the beginning of 2015 was $3.24 trillion.

As per the given data
Similarly no of students who got marks 30 and above = 44
Similarly no of students who got marks 40 and above = 25
∴ no of students whose marks are between 30 and 40 = 44 – 25 = 19 

From the line graph,
Ratio of income to expenditure in the 2003 = 2
Let the expenditure be a, ∴ income = 2a
Profit = income – expenditure
⇒ 450000 = 2a – a
⇒ a = Rs. 450000
Thus, income = 2a = Rs. 900000

The line graph is drawn between ratio of income to expenditure and annual year.
We don’t know the income in each year.
Thus, the data given is inadequate.